Oblique Pyramids (1) Can we have a rectangle cross-section (doesn't need to be parallel to the base) in an oblique square pyramid?
(2) Can we have a square cross-section (doesn't need to be parallel to the base) in an oblique rectangle pyramid? 
 A: Answer to question $1$:
Without loss of generality, let the oblique square pyramid $VABCD$ such that $A=(1,0,0)$, $B=(0,1,0)$, $C=(-1,0,0)$, $D=(0,-1,0)$ and $V=(e,f,g)$, where $e^2 +f^2 \neq 0$ and $g \neq 0$.
Let the points $M$, $N$, $P$ and $Q$ such that $M$ is in the line defined by $VA$, $N$ is in the line defined by $VB$, $P$ is in the line defined by $VC$ and $Q$ is in the line defined by $VD$.
If $MNPQ$ is a rectangle than:
$$\overrightarrow{NM}=\overrightarrow{PQ}.\quad (1)$$
Let $\lambda$, $\mu$, $\rho$ and $\nu$, real numbers, such that:
$$M=V+ \lambda \overrightarrow{VA},\quad (2)$$
$$N=V+ \mu \overrightarrow{VB},\quad (3)$$
$$P=V+ \rho \overrightarrow{VC} \quad (4)$$
and
$$Q=V+ \nu \overrightarrow{VD}. \quad (5)$$
Substituting $(2)$, $(3)$, $(4)$ and $(5)$ in $(1)$ we get three equations:
$$\lambda- \lambda e + \mu e = - \nu e + \rho + \rho e \quad (6) $$
$$-\lambda f - \mu + \mu f = - \nu - \nu f  + \rho f \quad (7) $$
$$-\lambda g + \mu g = - \nu g + \rho g . \quad (8) $$
Solving $(6)$, $(7)$ and $(8)$ we get:
$$\lambda = \mu = \nu = \rho. \quad(9)$$
Conclusion: the only way to get a rectangle is getting another square parallel to the pyramid base.
